Question

The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does the volume of the balloon change?

Answer 1

Answer:

6.1%

Explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law  at constant pressure,

$\frac{V_1}{T_1} =\frac{V_2}{T_2}$

$T_2=39\°\ C=39+273=312\ K$

$T_1=59°\ C=273+59=332\ K$

$V_2=V_1\times \frac{T_2}{T_1} \\=V_1\times\frac{312}{332}\\=0.939V_1$

Percentage change in volume

$\%\ change\ in\ volume=\frac{V_1-0.939V_1}{V_1} \times 100\\=6.1\%$

Change in volume of the balloon is 6.1%